Now look at the audion characteristic. Making the grid positive, that is going on the positive side of the zero volts in our map, makes the plate current larger. You remember that I told you in Letter 6 how the grid, when positive, helped call electrons away from the filament and so made a larger stream of electrons in the plate circuit. The grid calls electrons away from the filament. It can't call them out of it; they have to come out themselves as I explained to you in the fifth letter. [Illustration: Fig 23] You can see that as we make the grid more and more positive, that is, make it call louder and louder, a condition will be reached where it won't do it any good to call any louder, for it will already be getting all the electrons away from the filament just as fast as they are emitted. Making the grid more positive after that will not increase the plate current any. That's why the characteristic flattens off as you see at high values of grid voltage. The arrangement which we pictured in Fig. 22 for making changes in the grid voltage is simple but it doesn't let us change the voltage by less than that of a single battery cell. I want to show you a way which will. You'll find it very useful to know and it is easily understood for it is something like the arrangement of Fig. 14 in the preceding letter. [Illustration: Fig 24] Connect the cells as in Fig. 24 to a fine wire. About the middle of this wire connect the filament. As before use a clip on the end of the wire from the grid. If the grid is connected to _a_ in the figure there is applied to the grid circuit that part of the e. m. f. of the battery which is active in the length of wire between _o_ and _a_. The point _a_ is nearer the positive plate of the battery than is the point _o_. So the grid will be positive and the filament negative. On the other hand, if the clip is connected at _b_ the grid will be negative with respect to the filament. We can, therefore, make the grid positive or negative depending on which side of _o_ we connect the clip. How large the e. m. f. is which will be applied to the grid depends, of course, upon how far away from _o_ the clip is connected. Suppose you took the clip in your hand and slid it along in contact with the wire, first from _o_ to _a_ and then back again through _o_ to _b_ and so on back and forth. You would be making the grid _alternately_ positive and negative, wouldn't you? That is, you would be applying to the grid an e. m. f. which increases to some positive value and then, decreasing to zero, _reverses_, and increases just as much, only to decrease to zero, where it started. If you do this over and over again, taking always the same time for one round trip of the clip you will be impressing on the grid circuit an "_alternating e. m. f._"